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Popular >

cos^2(4x)-sin^2(4x)=cos(b)

Solution

cos^{2} (4 x) - sin^{2} (4 x) = cos (b)

Solution

x = \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}, x = - \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}

Solution steps

cos^{2} (4 x) - sin^{2} (4 x) = cos (b)

Rewrite using trig identities

cos^{2} (4 x) - sin^{2} (4 x)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

= cos (2 \cdot 4 x)

cos (2 \cdot 4 x) = cos (b)

Apply trig inverse properties

cos (2 \cdot 4 x) = cos (b)

General solutions for

cos (2 \cdot 4 x) = cos (b)

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

2 \cdot 4 x = arccos (cos (b)) + 2 πn, 2 \cdot 4 x = - arccos (cos (b)) + 2 πn

2 \cdot 4 x = arccos (cos (b)) + 2 πn, 2 \cdot 4 x = - arccos (cos (b)) + 2 πn

Solve

2 \cdot 4 x = arccos (cos (b)) + 2 πn : x = \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}

2 \cdot 4 x = arccos (cos (b)) + 2 πn

Divide both sides by

8

2 \cdot 4 x = arccos (cos (b)) + 2 πn

Divide both sides by

8

\frac{2 \cdot 4 x}{8} = \frac{arccos ( cos ( b ) )}{8} + \frac{2 πn}{8}

Simplify

x = \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}

x = \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}

Solve

2 \cdot 4 x = - arccos (cos (b)) + 2 πn : x = - \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}

2 \cdot 4 x = - arccos (cos (b)) + 2 πn

Divide both sides by

8

2 \cdot 4 x = - arccos (cos (b)) + 2 πn

Divide both sides by

8

\frac{2 \cdot 4 x}{8} = - \frac{arccos ( cos ( b ) )}{8} + \frac{2 πn}{8}

Simplify

x = - \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}

x = - \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}

x = \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}, x = - \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}

Graph

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